XSLT: output list on condition of attribute of parent-node
I'm struggling to create a list to be displayed on the condition of the parent node ('folder') with attribute 'folded' set to either 'yes' or 'no'. The result shall display the first two levels of the list only and not the third level as belo开发者_开发知识库w.
- 1st. Level: display
- 2nd. Level: display
- 3rd. Level: NO display
The idea is to check the 'folder'-attribute <folder folded="yes"> with <xsl:if test="not(parent::yes)">.
That should qualify for the 3rd. Level to NOT being displayed, but somehow it doesn't do what I want it to do.
I probably use the wrong construct and/or syntax.
Assistance is highly appreciated, thanks.
The XML structure with some content:
<xbel>
<folder folded="yes">
<level>1</level>
<title>bookmarks</title>
<desc>my bookmarks</desc>
<folder folded="no">
<level>2</level>
<title>Android</title>
<desc>my Android</desc>
<bookmark href="http://www.phonesreview.co.uk/">
<title>HTC Sync 3.0.5422 Update: Aria, Desire, Hero, Legend</title>
<desc>The new HTC Sync 3.0.5422 update will be most welcome...</desc>
</bookmark>
<folder folded="no">
<level>3</level>
<title>Apps</title>
<desc>Android Apps</desc>
<bookmark href="http://www.androidzoom.com/">
<title>Android Communication Apps</title>
<desc>Download Communication Apps for Android.</desc>
</bookmark>
<bookmark href="http://www.htc.com/">
<title>HTC - Android</title>
<desc>Apps for HTC-Android.</desc>
</bookmark>
</folder>
</folder>
</folder>
</xbel>
The XSLT:
<!--creates a nested list of elements named 'folder'-->
<xsl:template match="folder" mode="linklist">
<li>
<xsl:if test="folder/level = 2">
Level:<xsl:value-of select="level"/> /
Title:<xsl:value-of select="title"/> /
Desc:<xsl:value-of select="desc"/>
<ul>
<xsl:apply-templates mode="linklist" />
</ul>
</xsl:if>
</li>
</xsl:template>
<xsl:template match="bookmark" mode="linklist">
<li> <!-- this bookmark is just another item in the list of bookmarks -->
<!-- the title -->
<a rel="nofollow" href="{@href}"><xsl:value-of select="title"/></a>
<!-- the description -->
<xsl:if test="desc">
<span class="bookmarkDesc">
<xsl:value-of select="desc"/>
</span>
</xsl:if>
</li>
</xsl:template>
The Stylesheet HTML
<body>
<ul>
<xsl:apply-templates mode="linklist" />
</ul>
</body>
The generated output (levels:1-3)
Level:1 / Title:bookmarks / Desc:my bookmarks
Level:2 / Title:Android / Desc:my Android
HTC Sync 3.0.5422 Update: Aria, Desire, Hero, Legend ...
Level:3 / Title:Apps / Desc:Android Apps
Android Communication AppsDownload Communication Apps for Android.
HTC - AndroidApps for HTC-Android.
The anticipated output: (levels: 1-2)
Level:1 / Title:bookmarks / Desc:my bookmarks
Level:2 / Title:Android / Desc:my Android
HTC Sync 3.0.5422 Update: Aria, Desire, Hero, Legend ...
I tried this template, but that outputs the last two nodes, I need the two first nodes.
<xsl:template match="folder[parent::folder/@folded = 'yes']" mode="linklist">
The simplest possible change you could make to prevent processing of unfolded folder elements is to add an empty template that swallows them (i.e. produces no output):
<xsl:template match="folder[@folded='no']" mode="linklist"/>
All folder elements not having a folded attribute equal to no will be processed using your existing template; those that do will be captured by this new one.
If instead you want to process each folder element having either its own folded attribute equal to yes or that of its parent (as in the updated XML example), then use the following template:
<xsl:template match="folder[@folded='yes' or ../@folded='yes']" mode="linklist">
<!-- body elided -->
</xsl:template>
You'll probably also want to include an empty template for hiding all other folder elements:
<xsl:template match="folder" mode="linklist" />
加载中,请稍侯......
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